\(\int \frac {(a+b x+c x^2)^2}{(d+e x)^5} \, dx\) [2127]

Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {\left (c d^2-b d e+a e^2\right )^2}{4 e^5 (d+e x)^4}+\frac {2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{3 e^5 (d+e x)^3}-\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{2 e^5 (d+e x)^2}+\frac {2 c (2 c d-b e)}{e^5 (d+e x)}+\frac {c^2 \log (d+e x)}{e^5} \]

[Out]

-1/4*(a*e^2-b*d*e+c*d^2)^2/e^5/(e*x+d)^4+2/3*(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)/e^5/(e*x+d)^3+1/2*(-6*c^2*d^2-b^
2*e^2+2*c*e*(-a*e+3*b*d))/e^5/(e*x+d)^2+2*c*(-b*e+2*c*d)/e^5/(e*x+d)+c^2*ln(e*x+d)/e^5

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {712} \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{2 e^5 (d+e x)^2}+\frac {2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{3 e^5 (d+e x)^3}-\frac {\left (a e^2-b d e+c d^2\right )^2}{4 e^5 (d+e x)^4}+\frac {2 c (2 c d-b e)}{e^5 (d+e x)}+\frac {c^2 \log (d+e x)}{e^5} \]

[In]

Int[(a + b*x + c*x^2)^2/(d + e*x)^5,x]

[Out]

-1/4*(c*d^2 - b*d*e + a*e^2)^2/(e^5*(d + e*x)^4) + (2*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(3*e^5*(d + e*x)^
3) - (6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))/(2*e^5*(d + e*x)^2) + (2*c*(2*c*d - b*e))/(e^5*(d + e*x)) + (
c^2*Log[d + e*x])/e^5

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (c d^2-b d e+a e^2\right )^2}{e^4 (d+e x)^5}+\frac {2 (-2 c d+b e) \left (c d^2-b d e+a e^2\right )}{e^4 (d+e x)^4}+\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^4 (d+e x)^3}-\frac {2 c (2 c d-b e)}{e^4 (d+e x)^2}+\frac {c^2}{e^4 (d+e x)}\right ) \, dx \\ & = -\frac {\left (c d^2-b d e+a e^2\right )^2}{4 e^5 (d+e x)^4}+\frac {2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{3 e^5 (d+e x)^3}-\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{2 e^5 (d+e x)^2}+\frac {2 c (2 c d-b e)}{e^5 (d+e x)}+\frac {c^2 \log (d+e x)}{e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^2 d \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )-e^2 \left (3 a^2 e^2+2 a b e (d+4 e x)+b^2 \left (d^2+4 d e x+6 e^2 x^2\right )\right )-2 c e \left (a e \left (d^2+4 d e x+6 e^2 x^2\right )+3 b \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )+12 c^2 (d+e x)^4 \log (d+e x)}{12 e^5 (d+e x)^4} \]

[In]

Integrate[(a + b*x + c*x^2)^2/(d + e*x)^5,x]

[Out]

(c^2*d*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3*x^3) - e^2*(3*a^2*e^2 + 2*a*b*e*(d + 4*e*x) + b^2*(d^2 +
4*d*e*x + 6*e^2*x^2)) - 2*c*e*(a*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + 3*b*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^
3)) + 12*c^2*(d + e*x)^4*Log[d + e*x])/(12*e^5*(d + e*x)^4)

Maple [A] (verified)

Time = 3.00 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.21

method	result	size
risch	\(\frac {-\frac {2 c \left (b e -2 c d \right ) x^{3}}{e^{2}}-\frac {\left (2 a c \,e^{2}+b^{2} e^{2}+6 b c d e -18 c^{2} d^{2}\right ) x^{2}}{2 e^{3}}-\frac {\left (2 a b \,e^{3}+2 d \,e^{2} a c +b^{2} d \,e^{2}+6 b c e \,d^{2}-22 c^{2} d^{3}\right ) x}{3 e^{4}}-\frac {3 a^{2} e^{4}+2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}+6 d^{3} e b c -25 c^{2} d^{4}}{12 e^{5}}}{\left (e x +d \right )^{4}}+\frac {c^{2} \ln \left (e x +d \right )}{e^{5}}\)	\(182\)
norman	\(\frac {-\frac {3 a^{2} e^{4}+2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}+6 d^{3} e b c -25 c^{2} d^{4}}{12 e^{5}}-\frac {2 \left (b c e -2 c^{2} d \right ) x^{3}}{e^{2}}-\frac {\left (2 a c \,e^{2}+b^{2} e^{2}+6 b c d e -18 c^{2} d^{2}\right ) x^{2}}{2 e^{3}}-\frac {\left (2 a b \,e^{3}+2 d \,e^{2} a c +b^{2} d \,e^{2}+6 b c e \,d^{2}-22 c^{2} d^{3}\right ) x}{3 e^{4}}}{\left (e x +d \right )^{4}}+\frac {c^{2} \ln \left (e x +d \right )}{e^{5}}\)	\(184\)
default	\(-\frac {2 c \left (b e -2 c d \right )}{e^{5} \left (e x +d \right )}-\frac {2 a b \,e^{3}-4 d \,e^{2} a c -2 b^{2} d \,e^{2}+6 b c e \,d^{2}-4 c^{2} d^{3}}{3 e^{5} \left (e x +d \right )^{3}}-\frac {a^{2} e^{4}-2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-2 d^{3} e b c +c^{2} d^{4}}{4 e^{5} \left (e x +d \right )^{4}}-\frac {2 a c \,e^{2}+b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}}{2 e^{5} \left (e x +d \right )^{2}}+\frac {c^{2} \ln \left (e x +d \right )}{e^{5}}\)	\(193\)
parallelrisch	\(\frac {-12 x^{2} a c \,e^{4}-2 a b d \,e^{3}-b^{2} d^{2} e^{2}-3 a^{2} e^{4}+12 \ln \left (e x +d \right ) c^{2} d^{4}+108 x^{2} c^{2} d^{2} e^{2}+88 x \,c^{2} d^{3} e +25 c^{2} d^{4}+48 \ln \left (e x +d \right ) x \,c^{2} d^{3} e -6 d^{3} e b c +48 x^{3} c^{2} d \,e^{3}-8 x a c d \,e^{3}-6 x^{2} b^{2} e^{4}+12 \ln \left (e x +d \right ) x^{4} c^{2} e^{4}-24 x^{3} b c \,e^{4}-8 x a b \,e^{4}-4 x \,b^{2} d \,e^{3}-36 x^{2} b c d \,e^{3}-24 x b c \,d^{2} e^{2}+72 \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{2}+48 \ln \left (e x +d \right ) x^{3} c^{2} d \,e^{3}-2 a c \,d^{2} e^{2}}{12 e^{5} \left (e x +d \right )^{4}}\)	\(268\)

[In]

int((c*x^2+b*x+a)^2/(e*x+d)^5,x,method=_RETURNVERBOSE)

[Out]

(-2*c*(b*e-2*c*d)/e^2*x^3-1/2*(2*a*c*e^2+b^2*e^2+6*b*c*d*e-18*c^2*d^2)/e^3*x^2-1/3*(2*a*b*e^3+2*a*c*d*e^2+b^2*
d*e^2+6*b*c*d^2*e-22*c^2*d^3)/e^4*x-1/12*(3*a^2*e^4+2*a*b*d*e^3+2*a*c*d^2*e^2+b^2*d^2*e^2+6*b*c*d^3*e-25*c^2*d
^4)/e^5)/(e*x+d)^4+c^2*ln(e*x+d)/e^5

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.75 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {25 \, c^{2} d^{4} - 6 \, b c d^{3} e - 2 \, a b d e^{3} - 3 \, a^{2} e^{4} - {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 24 \, {\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x^{3} + 6 \, {\left (18 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} - {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 4 \, {\left (22 \, c^{2} d^{3} e - 6 \, b c d^{2} e^{2} - 2 \, a b e^{4} - {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x + 12 \, {\left (c^{2} e^{4} x^{4} + 4 \, c^{2} d e^{3} x^{3} + 6 \, c^{2} d^{2} e^{2} x^{2} + 4 \, c^{2} d^{3} e x + c^{2} d^{4}\right )} \log \left (e x + d\right )}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} \]

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^5,x, algorithm="fricas")

[Out]

1/12*(25*c^2*d^4 - 6*b*c*d^3*e - 2*a*b*d*e^3 - 3*a^2*e^4 - (b^2 + 2*a*c)*d^2*e^2 + 24*(2*c^2*d*e^3 - b*c*e^4)*
x^3 + 6*(18*c^2*d^2*e^2 - 6*b*c*d*e^3 - (b^2 + 2*a*c)*e^4)*x^2 + 4*(22*c^2*d^3*e - 6*b*c*d^2*e^2 - 2*a*b*e^4 -
(b^2 + 2*a*c)*d*e^3)*x + 12*(c^2*e^4*x^4 + 4*c^2*d*e^3*x^3 + 6*c^2*d^2*e^2*x^2 + 4*c^2*d^3*e*x + c^2*d^4)*log
(e*x + d))/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d^4*e^5)

Sympy [A] (verification not implemented)

Time = 17.03 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^{2} \log {\left (d + e x \right )}}{e^{5}} + \frac {- 3 a^{2} e^{4} - 2 a b d e^{3} - 2 a c d^{2} e^{2} - b^{2} d^{2} e^{2} - 6 b c d^{3} e + 25 c^{2} d^{4} + x^{3} \left (- 24 b c e^{4} + 48 c^{2} d e^{3}\right ) + x^{2} \left (- 12 a c e^{4} - 6 b^{2} e^{4} - 36 b c d e^{3} + 108 c^{2} d^{2} e^{2}\right ) + x \left (- 8 a b e^{4} - 8 a c d e^{3} - 4 b^{2} d e^{3} - 24 b c d^{2} e^{2} + 88 c^{2} d^{3} e\right )}{12 d^{4} e^{5} + 48 d^{3} e^{6} x + 72 d^{2} e^{7} x^{2} + 48 d e^{8} x^{3} + 12 e^{9} x^{4}} \]

[In]

integrate((c*x**2+b*x+a)**2/(e*x+d)**5,x)

[Out]

c**2*log(d + e*x)/e**5 + (-3*a**2*e**4 - 2*a*b*d*e**3 - 2*a*c*d**2*e**2 - b**2*d**2*e**2 - 6*b*c*d**3*e + 25*c
**2*d**4 + x**3*(-24*b*c*e**4 + 48*c**2*d*e**3) + x**2*(-12*a*c*e**4 - 6*b**2*e**4 - 36*b*c*d*e**3 + 108*c**2*
d**2*e**2) + x*(-8*a*b*e**4 - 8*a*c*d*e**3 - 4*b**2*d*e**3 - 24*b*c*d**2*e**2 + 88*c**2*d**3*e))/(12*d**4*e**5
+ 48*d**3*e**6*x + 72*d**2*e**7*x**2 + 48*d*e**8*x**3 + 12*e**9*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {25 \, c^{2} d^{4} - 6 \, b c d^{3} e - 2 \, a b d e^{3} - 3 \, a^{2} e^{4} - {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 24 \, {\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x^{3} + 6 \, {\left (18 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} - {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 4 \, {\left (22 \, c^{2} d^{3} e - 6 \, b c d^{2} e^{2} - 2 \, a b e^{4} - {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} + \frac {c^{2} \log \left (e x + d\right )}{e^{5}} \]

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^5,x, algorithm="maxima")

[Out]

1/12*(25*c^2*d^4 - 6*b*c*d^3*e - 2*a*b*d*e^3 - 3*a^2*e^4 - (b^2 + 2*a*c)*d^2*e^2 + 24*(2*c^2*d*e^3 - b*c*e^4)*
x^3 + 6*(18*c^2*d^2*e^2 - 6*b*c*d*e^3 - (b^2 + 2*a*c)*e^4)*x^2 + 4*(22*c^2*d^3*e - 6*b*c*d^2*e^2 - 2*a*b*e^4 -
(b^2 + 2*a*c)*d*e^3)*x)/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d^4*e^5) + c^2*log(e*x + d)/e^
5

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (144) = 288\).

Time = 0.27 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.04 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {c^{2} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{5}} + \frac {\frac {48 \, c^{2} d e^{15}}{e x + d} - \frac {36 \, c^{2} d^{2} e^{15}}{{\left (e x + d\right )}^{2}} + \frac {16 \, c^{2} d^{3} e^{15}}{{\left (e x + d\right )}^{3}} - \frac {3 \, c^{2} d^{4} e^{15}}{{\left (e x + d\right )}^{4}} - \frac {24 \, b c e^{16}}{e x + d} + \frac {36 \, b c d e^{16}}{{\left (e x + d\right )}^{2}} - \frac {24 \, b c d^{2} e^{16}}{{\left (e x + d\right )}^{3}} + \frac {6 \, b c d^{3} e^{16}}{{\left (e x + d\right )}^{4}} - \frac {6 \, b^{2} e^{17}}{{\left (e x + d\right )}^{2}} - \frac {12 \, a c e^{17}}{{\left (e x + d\right )}^{2}} + \frac {8 \, b^{2} d e^{17}}{{\left (e x + d\right )}^{3}} + \frac {16 \, a c d e^{17}}{{\left (e x + d\right )}^{3}} - \frac {3 \, b^{2} d^{2} e^{17}}{{\left (e x + d\right )}^{4}} - \frac {6 \, a c d^{2} e^{17}}{{\left (e x + d\right )}^{4}} - \frac {8 \, a b e^{18}}{{\left (e x + d\right )}^{3}} + \frac {6 \, a b d e^{18}}{{\left (e x + d\right )}^{4}} - \frac {3 \, a^{2} e^{19}}{{\left (e x + d\right )}^{4}}}{12 \, e^{20}} \]

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^5,x, algorithm="giac")

[Out]

-c^2*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^5 + 1/12*(48*c^2*d*e^15/(e*x + d) - 36*c^2*d^2*e^15/(e*x + d)^2
+ 16*c^2*d^3*e^15/(e*x + d)^3 - 3*c^2*d^4*e^15/(e*x + d)^4 - 24*b*c*e^16/(e*x + d) + 36*b*c*d*e^16/(e*x + d)^2
- 24*b*c*d^2*e^16/(e*x + d)^3 + 6*b*c*d^3*e^16/(e*x + d)^4 - 6*b^2*e^17/(e*x + d)^2 - 12*a*c*e^17/(e*x + d)^2
+ 8*b^2*d*e^17/(e*x + d)^3 + 16*a*c*d*e^17/(e*x + d)^3 - 3*b^2*d^2*e^17/(e*x + d)^4 - 6*a*c*d^2*e^17/(e*x + d
)^4 - 8*a*b*e^18/(e*x + d)^3 + 6*a*b*d*e^18/(e*x + d)^4 - 3*a^2*e^19/(e*x + d)^4)/e^20

Mupad [B] (verification not implemented)

Time = 9.85 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^2\,\ln \left (d+e\,x\right )}{e^5}-\frac {x^2\,\left (\frac {b^2\,e^4}{2}+3\,b\,c\,d\,e^3-9\,c^2\,d^2\,e^2+a\,c\,e^4\right )+x\,\left (\frac {b^2\,d\,e^3}{3}+2\,b\,c\,d^2\,e^2+\frac {2\,a\,b\,e^4}{3}-\frac {22\,c^2\,d^3\,e}{3}+\frac {2\,a\,c\,d\,e^3}{3}\right )-x^3\,\left (4\,c^2\,d\,e^3-2\,b\,c\,e^4\right )+\frac {a^2\,e^4}{4}-\frac {25\,c^2\,d^4}{12}+\frac {b^2\,d^2\,e^2}{12}+\frac {a\,b\,d\,e^3}{6}+\frac {b\,c\,d^3\,e}{2}+\frac {a\,c\,d^2\,e^2}{6}}{e^5\,{\left (d+e\,x\right )}^4} \]

[In]

int((a + b*x + c*x^2)^2/(d + e*x)^5,x)

[Out]

(c^2*log(d + e*x))/e^5 - (x^2*((b^2*e^4)/2 - 9*c^2*d^2*e^2 + a*c*e^4 + 3*b*c*d*e^3) + x*((b^2*d*e^3)/3 - (22*c
^2*d^3*e)/3 + (2*a*b*e^4)/3 + (2*a*c*d*e^3)/3 + 2*b*c*d^2*e^2) - x^3*(4*c^2*d*e^3 - 2*b*c*e^4) + (a^2*e^4)/4 -
(25*c^2*d^4)/12 + (b^2*d^2*e^2)/12 + (a*b*d*e^3)/6 + (b*c*d^3*e)/2 + (a*c*d^2*e^2)/6)/(e^5*(d + e*x)^4)